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2 years ago

A hazard sign has 3 identical

1 answer:

5 0

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

$h=\sqrt{18^{2}+6^{2}}$

$h=\sqrt{360}$

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

$P_{t}=$ 44 + 44 + 44

$P_{t}=$ 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

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The x coordinate of an electron is measured with an uncertainty of 0.200 mm . What is vx, the x component of the electron's velo

Vikentia [17]

Answer:

Velocity of electron along x direction is 57.9 m/s

Explanation:

The uncertainty in x coordinate of electron, Δx = 0.200 mm = 0.2 x 10⁻³ m

Let vₓ be the x component of electrons velocity.

The uncertainty in x component of electrons momentum is:

Δpₓ = mΔvₓ

Here m is mass of the electron.

The uncertainty in velocity x component is 1% i.e. 0.01.

So, the above equation can be written as :

Δpₓ = 0.01mvₓ ....(1)

The minimum uncertainty principle is:

$\Delta x\Delta p_{x} = \frac{h}{2\pi }$ ....(2)

Here h is Planck's constant.

From equation (1) and (2),

$\Delta x\times0.01m v_{x} = \frac{h}{2\pi }$

Substitute 0.2 x 10⁻³ m for Δx, 9.1 x 10⁻³¹ kg for m and 6.626 x 10⁻³⁴ m²kg/s in the above equation.

$0.2\times10^{-3} \times0.01\times9.1\times10^{-31}\times v_{x} = \frac{6.626\times10^{-34} }{2\pi }$

vₓ = 57.9 m/s

8 0

2 years ago

Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

5 0

2 years ago

A 12.0 V car battery is being used to power the headlights of a car. Each of the two headlights has a power rating of 41.8 Watts

Marina CMI [18]

Answer:

n = 4.35 x 10¹⁹

Explanation:

Given that

Voltage V= 12 V

Power rating of headlights = 41.8 W

We know that headlights of the car is connected in parallel connection.

We know that

Power = Current x Voltage

P = V I

$I=\dfrac{41.8}{12}\ A$

I=3.48 A

Therefore the total current will be

I'= 3.48 + 3.48 A

I = 6.96 A

We know that

Charge = Current x time

q= I' t

q= 6.96 x 1

q= 6.96 C

The charge on electron ,e= 1.6 x 10⁻¹⁹

q= n e

n=Number of electron

$n=\dfrac{q}{e}$

$n=\dfrac{6.96}{1.6\times 10^{-19}}$

n = 4.35 x 10¹⁹

5 0

3 years ago

A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an

Roman55 [17]

Answer:

$v_{2f} = \frac{2vm_1}{m_2 + m_1}$

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

$P_i = m_1v$

After the collision

$P_f = m_1v_{1f} + m_2v_{2f}$

So using the law of momentum conservation

$P_i = P_f$

$m_1v = m_1v_{1f} + m_2v_{2f}$

We can solve for the speed of ball 1 post collision in terms of others:

$v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

Their kinetic energy is also conserved before and after collision

$m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2$

$m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2$

From here we can plug in $v_{1f} = v - v_{2f}\frac{m_2}{m_1}$

$m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2$

$m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2$

$m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2$

$v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0$

$v_{2f}(1 + \frac{m_2}{m_1}) = 2v$

$v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}$

8 0

3 years ago

Read 2 more answers

A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density? 0.081 kg/m3 12.29 g/cm3 132912.00 g/cm3 canno

REY [17]

The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.

4 0

2 years ago

Read 2 more answers